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frez [133]
2 years ago
6

A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement? 4 miles 16 miles 20 miles 2

8 miles
Physics
2 answers:
Ostrovityanka [42]2 years ago
7 0
For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
 We have then
 d ^ 2 = 16 ^ 2 + 12 ^ 2

 From here, we clear the value of d.
 We have then:
 d =  \sqrt{16 ^ 2 + 12 ^ 2} 

 Rewriting:
 d = \sqrt{256 + 144}
 d = \sqrt{400}
 d = 20 miles

 Answer:
 
The magnitude of the car's displacement is:
 
d = 20 miles
Lilit [14]2 years ago
3 0
20 miles. just took the quiz.
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Sir Marvin decided to improve the destructive power of his cannon by increasing the size of his cannonballs. Sir Seymour kept hi
maria [59]
We really can't tell from the given information. 
We don't know HOW MUCH Marv enlarged his cannonballs,
or HOW MUCH faster Seymour's balls became.

If we assume that they both, let's say, DOUBLED something,
then Seymour accomplished more, and the destructive capability
of his balls has increased more. 

I say that because the destructive capability of a cannonball is
pretty much just its kinetic energy when it arrives and hits the target.
Now, we all know the equation for kinetic energy.

                K.E.  =  (1/2) (mass) (speed-SQUARED) .

We can see right away that if Marv started shooting balls with
double the mass but at the same speed, then they have double
the kinetic energy of the old ones.

But if Seymour started shooting the same balls with double the SPEED,
then they have (2-SQUARED) as much kinetic energy as they used to.

That's 4 times as much destructive capability as before.  

So we can say that when it comes to cannons and their balls and
smashing things to bits and terrorizing your opponents, if making
a bigger mess is better, then more mass is better, but more speed
is better-squared.
5 0
2 years ago
A moving 46.6 kg sled feels a 52.9 N friction force. what is the coefficient of friction
Setler [38]

Answer:

F=UR

52.9=U*46.6

U=52.9/46.6

U=1.135

4 0
2 years ago
Read 2 more answers
If no friction acts on a diver during a dive, then which of the following statements is true? A) The total mechanical energy of
EleoNora [17]
If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.

Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).

The initial PE = mgh₁ and the initial KE  = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
 
Therefore
(KE + PE) ₁ = (KE + PE)₂

Evaluate the given answers.
A) The total mechanical energy of the system increases.
     FALSE

B) Potential energy can be converted into kinetic energy but not vice versa.
     TRUE

C) (KE + PE)beginning = (KE + PE) end.
     TRUE

D) All of the above.
     FALSE

4 0
2 years ago
Read 2 more answers
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
1 year ago
A roadway for stunt drivers is designed for racecars moving at a speed of 40 m/s. A curved section of the roadway is a circular
Fynjy0 [20]

Answer:

Bank angle = 35.34o

Explanation:

Since the road is frictionless,

Tan (bank angle) = V^2/r*g

Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2

Tan ( bank angle) = 40^2/(230*9.81)

Tan (bank angle) = 0.7091

Bank angle = tan inverse (0.7091)

Bank angle = 35.34o

3 0
1 year ago
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